Joint density for X and Y is
$f(x,y)=2e^{-x-2y} \ \ \ \ \ \ $for$ \ \ \ x>0,y>0 $
$0$ otherwise
Calculate the variance of $Y$ given that $X>3$ and $Y>3$
Can we use this formula somehow here $V(Y)=E(V(X|Y))+V(E(X|Y))$?
I thought that because joint density seems like product of two exponential distributions one can be conditional density and other can be marginal. Am i think totally wrong ? Bad way to deal this problem i guess. When i tried to solve this it was very laborious.
$X$ and $Y$ are obviously independent so we can calculate the variance of $Y|x>3,Y>3$ easily. First note that $$Y|x>3,Y>3=Y|Y>3$$ so the variance is$$Var[Y|Y>3]=E[Y^2|Y>3]-E^2[Y|Y>3]$$also knowing $Pr(Y>3)=e^{-6}$ we have: $$E[Y^2|Y>3]=\dfrac{1}{e^{-6}}\int_{3}^{\infty}2y^2e^{-2y}dy=\dfrac{25}{2}$$and $$E[Y|Y>3]=\dfrac{1}{e^{-6}}\int_{3}^{\infty}2ye^{-2y}dy=\dfrac{7}{2}$$therefore$$Var[Y|Y>3]=\dfrac{25}{2}-\dfrac{49}{4}=\dfrac{1}{4}$$